X, Y are independent normal random variables. What is the pdf of Z = X/(X+Y)?

Question

This is a question on one of final exam at my school...:

X and Y are independent Gaussian random variables with zero mean and unit variance. Z=X/(X+Y). Find the pdf of Z.

After some struggles, I think I can solve this question by letting W = X+Y first and derive the pdf of W. Then I can find the pdf of Z = X/W.

However, after reading these two wikipedia articles:

https://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

https://en.wikipedia.org/wiki/Ratio_distribution.

I realized that the whole answer is going to be very long.

Meanwhile, I came across this posting - Finding the pdf of $(X+Y)^2/(X^2+Y^2)$ where $X$ and $Y$ are independent and normal, which is similar in nature.

It appears that what are shown on the wikipedia articles are the only way to go.

However, considering this is a final exam question, given a short time limit, there should probably a much shorter way to reach my answer. Can anyone please show me the "easier" way?

Answer 1

We know that $$\frac{X}{X+Y}=1-\frac{Y}{X+Y}.$$ Note that the last term is simply the equivalent of the first in $Y$. Can you now solve the problem, given this insight?

Also, just to correct a small misconception; for us to calculate the ratio distribution, the variables must be independent. With your proposed method $(Z=X+Y)$, they are not, so we cannot use that method.

Answer 2

I think I had missed the given conditions that $\mu_{x}=\mu_{y}=0$ and that $\sigma_{x}=\sigma_{y}=1$.

I realized if I take these into account, I will come out with

$f_{W}(w)=\frac{1}{\sqrt{2\pi }}e^{-\frac{w^{2}}{2}}$

for pdf of W.

Then, continue working with reference on the ratio distribution derivations I will have

$f_{Z}(z)=\frac{1}{\pi (z^{2}+1)}$

Therefore, the simplification comes if I had remembered to substitute in $\mu_{x}=\mu_{y}=0$ and that $\sigma_{x}=\sigma_{y}=1$